If $\mathop {\lim }\limits_{x \to \infty } {\left( {1 + \frac{a}{x} - \frac{4}{{{x^2}}}} \right)^{2x}} = {e^3},$ then $a$ is equal to

The value of $\lim_{x \rightarrow 1} \frac{(x^{2}-1) \sin^{2}(\pi x)}{x^{4}-2x^{3}+2x-1}$ is equal to

The value of $\lim _{x \rightarrow 0} \frac{x e^{x}-\sin x}{x}$ is equal to:

Find the limit: $\mathop {\lim }\limits_{x \to 2} \left[\frac{x^{2}-4}{x^{3}-4 x^{2}+4 x}\right]$

$\lim _{\theta \rightarrow \frac{\pi}{2}^{-}} \frac{8 \tan ^4 \theta+4 \tan ^2 \theta+5}{(3-2 \tan \theta)^4} = $

$\mathop {\lim }\limits_{x \to {a^ + }} \left( \frac{{|x{|^3}}}{a} - {\left[ {\frac{x}{a}} \right]^3} \right) \,(a > 0)$ is equal to :- (where $[x]$ is the greatest integer function and $|x|$ is the modulus function)